Kth Smallest Element in a BST

https://leetcode.com/problems/kth-smallest-element-in-a-bst/

Given a binary search tree, write a function

kthSmallest

to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?

Hint:

Try to utilize the property of a BST.

What if you could modify the BST node's structure?

The optimal runtime complexity is O(height of BST).

解题思路：

带返回值的递归

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int kthSmallest(TreeNode root, int k) {
int [] step = new int[1];
step[0] = 1;
TreeNode res = dfs(root, k, step);
if(res == null) {
return -1;
}
return res.val;
}

public TreeNode dfs(TreeNode root, int k, int[] step) {
if(root == null) {
return null;
}
TreeNode res1 = dfs(root.left, k, step);
if(res1 != null) {
return res1;
}
if(step[0] == k) {
return root;
}
step[0]++;
TreeNode res2 = dfs(root.right, k, step);
return res2;
}
}

不带返回值的递归

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int kthSmallest(TreeNode root, int k) {
int [] step = new int[1];
step[0] = 1;
int [] res = new int[1];
dfs(root, k, step, res);
return res[0];
}

public void dfs(TreeNode root, int k, int[] step, int[] res) {
if(root == null) {
return;
}
dfs(root.left, k, step, res);
if(step[0] == k) {
res[0] = root.val;
step[0]++;
return;
}
step[0]++;
dfs(root.right, k, step, res);
}
}

或者直接将step赋值为k也可以，省去一个参数

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int kthSmallest(TreeNode root, int k) {
int [] step = new int[1];
step[0] = k;
int [] res = new int[1];
dfs(root, step, res);
return res[0];
}

public void dfs(TreeNode root, int[] step, int[] res) {
if(root == null) {
return;
}
dfs(root.left, step, res);
if(step[0] == 1) {
res[0] = root.val;
step[0]--;
return;
}
step[0]--;
dfs(root.right, step, res);
}
}

参数使用数组的原因是Java无法pass by reference，或者使用全局变量，也是可以的。